Check if $ x-3 $ is a factor of $ x^{4}-2x^{3}-25x^{2}+26x+120 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-2&-25&26&120\\& & 3& 3& -66& \color{black}{-120} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-22}&\color{blue}{-40}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-3 $ is a factor of the $ x^{4}-2x^{3}-25x^{2}+26x+120 $.
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-2&-25&26&120\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-2&-25&26&120\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-2&-25&26&120\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 3 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -2 }&-25&26&120\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-2&-25&26&120\\& & 3& \color{blue}{3} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 3 } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}3&1&-2&\color{orangered}{ -25 }&26&120\\& & 3& \color{orangered}{3} & & \\ \hline &1&1&\color{orangered}{-22}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ -66 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-2&-25&26&120\\& & 3& 3& \color{blue}{-66} & \\ \hline &1&1&\color{blue}{-22}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -66 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}3&1&-2&-25&\color{orangered}{ 26 }&120\\& & 3& 3& \color{orangered}{-66} & \\ \hline &1&1&-22&\color{orangered}{-40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ -120 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-2&-25&26&120\\& & 3& 3& -66& \color{blue}{-120} \\ \hline &1&1&-22&\color{blue}{-40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 120 } + \color{orangered}{ \left( -120 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&1&-2&-25&26&\color{orangered}{ 120 }\\& & 3& 3& -66& \color{orangered}{-120} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-22}&\color{blue}{-40}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.